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A box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

A box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

Grade:11

1 Answers

Vikas TU
14149 Points
3 years ago
N = normal force by the incline on the block
f' = frictional force on the block
m = mass of the block = 8 kg
θ = angle of incline
μ = Coefficient of friction
consider the force diagram to prevent the motion of block in down direction :
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ    
parallel to incline , the force equation is given as
f + f' = mg Sinθ
f + μ mg Cosθ = mg Sinθ
f = mg Sinθ - μ mg Cosθ                                           eq-1
 
consider the force diagram to make the block slide upward:
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ    
parallel to incline , the force equation is given as
2 f = mg Sinθ + f'
2 f = mg Sinθ +  μ mg Cosθ
using eq-1
2 (mg Sinθ - μ mg Cosθ) = mg Sinθ +  μ mg Cosθ
2 mg Sinθ - 2 μ mg Cosθ = mg Sinθ +  μ mg Cosθ
mg Sinθ = 3  μ mg Cosθ
tanθ =  3  μ
μ = tanθ/3

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